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1 year ago

Tom’s company has been contracted to excavate uranium ore with minimal ground disruption. What process should his company use?

2 answers:

7 0

In-situ leaching or solution mining offers the least ground disruptive type of mining and waste. This type of mining only dissolves the uranium where it is under the ground then pump up to the ground and further processed through milling.

julia-pushkina [17]1 year ago

5 0

Answer:

In-Situ leaching

Explanation:

There are five ways by which Uranium is mined.

In-situ leach - 44.9%

Underground mining - 26.2%

Open pit - 19.9%

Heap leaching - 1.7%

7.3% was derived from processing of by products of other minerals

In-Situ leaching involves dissolving the minerals of the ore by pumping lixivant through boreholes made by drilling or Explosive or hydraulic fracturing. So, here not a lot of excavation is done just a hole is drilled or bored to the location of the ore, the ore is then made porous then lixivant is pumped. Then the solution is pumped out and further chemical processes are carried out to extract uranium from the solution.

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Uranus (mass = 8.68 x 1025 kg) and

Pepsi [2]

Answer:129,398,203.7 m

Explanation:

According to Newton's law of Universal Gravitation, the force $F$ exerted between two bodies of masses $M$ and $m$ and separated by a distance $d$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{Mm}{d^2}$ (1)

Where:

$F=2.28(10)^{19} N$ is the gravitational force

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$M=8.68(10)^{25} kg$ is the mass of Uranus

$m=6.59(10)^{19} kg$ is the mass of Uranu's moon, Mirana

$d$ is the distance between Uranus and its moon

Isolating $d$ :

$d=\sqrt{\frac{GMm}{F}}$ (2)

$d=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.68(10)^{25} kg)(6.59(10)^{19} kg)}{2.28(10)^{19} N}}$ (3)

Finally:

$d=129,398,203.7 m$

5 0

2 years ago

Whose research showed that atoms consist of small positively charged nuclear centers and lots of empty space populated by electr

Tju [1.3M]

Answer:

Rutherford

Explanation:

Basic principles of the Rutherford atomic model.

1. Positively charged particles are in a very small volume compared to the size of the atom.

2. Most of the mass of the atom is in that small central volume. Rutherford did not call it "core" in his initial papal but he did it from 1912.

3. Electrons with negative electrical charge revolve around the nucleus.

4. The electrons rotate at high speeds around the nucleus and in circular paths that it called orbits.

5. Both negatively charged electrons and the positively charged nucleus are held together by an electrostatic attraction force.

4 0

2 years ago

A counter attendant in a diner shoves a ketchup bottle with a mass 0.30 kg along a smooth, level lunch counter. The bottle leave

balu736 [363]

Answer:

1.176 N

Explanation:

$m$ = mass of the bottle = 0.30 kg

$v_{o}$ = initial speed of the bottle = 2.8 m/s

$v$ = final speed of the bottle = 0 m/s

$d$ = stopping distance traveled = 1.0 m

$f$ = magnitude of frictional force acting on bottle

Using work-change in kinetic energy theorem

$- f d = (0.5) m (v^{2} - v_{o}^{2} )\\- f (1) = (0.5) (0.30) (0^{2} - 2.8^{2} )\\-f = - 1.176 \\f = 1.176$ N

direction :

frictional force acts in opposite direction of motion.

8 0

2 years ago

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

Step 4: calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0

2 years ago

The drag force F on a boat varies jointly with the wet surface area A of the boat and the square of the speed s of the boat. A b

Advocard [28]

Answer:

Wet surfaces areaA=+25.3ft^2

Explanation:

Using F= K×A× S^2

Where F= drag force

A= surface area

S= speed

Given : F=996N S=20mph A= 83ft^2

K = F/AS^2=996/(83×20^2)

K= 996/33200 = 0.03

1215= (0.03)× A × 18^2

1215=9.7A

A=1215/9.7=125.3ft^2

7 0

2 years ago